Use the formula y=a(1+ rn)^(nt) where n is the number of times the initial value is compounded in a year.
C
c
Use the formula y=a(1+ rn)^(nt) where n is the number of times the initial value is compounded in a year.
D
d
Use the formula y=ae^(rt).
A
a
About 39.52 years.
B
b
About 38.66 years.
C
c
About 38.38 years.
D
d
About 38.38 years.
Practice makes perfect
a
The balance, y, can be modeled by an Exponential-Growth-Model.
y=a(1+r)^t
In this function, a is the initial amount, r is the percent increase expressed in decimal form, and t is the time in years. We know that the account starts with a= 100 and pays 6 % annual interest which, expressed in decimal form, can be written as r= 0.06.
y= 100(1+ 0.06)^t ⇔ y=100(1.06)^t
By setting y equal to 1000, we can solve for t.
To solve this equation, we need to use logarithms. Notice that the base of the power on the left-hand side is 1.06. Does this mean we have to use a logarithm with a base of 1.06? No, as long as the same logarithm is applied to both sides of the equation, we can use any logarithm we want.
If the frequency of compounding is quarterly, we have to use a formula that describes compound interest. Like in Part A, we already know that a= 100 and r= 0.06.
y=a(1+r/n)^(nt) ⇒ y= 100(1+0.06/n)^(nt)
In the formula, n is the number of times the initial principal is compounded per year. Since there are 4 quarters in a year, we have n=4
y=100(1+0.06/4)^(4t)
Like in Part A, we will set y equal to 1000 and solve for t.
When compounded daily, it will take about 38.38 years for the principal to reach $1000.
d
When interest is compounded continuously, we need to use a different formula for compounded interest.
y=ae^(rt)
Like in previous parts, a is the initial value and r is the annual interest rate expressed as a decimal.
y=100e^(0.06t)
By substituting y=1000, we can solve for t.
